Carnival ride, Friction and Normal Forces

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SUMMARY

The discussion focuses on the physics of a carnival ride where patrons experience centripetal force while standing against a cylindrical wall. The radius of the ride is 6.3 meters, and it spins at a frequency of 23.5 revolutions per minute. Key calculations include determining the normal force using the formula N = mv²/r and finding the minimum coefficient of friction required to prevent sliding. The engineers aim to ensure that the normal force does not exceed 1.7 times the weight of the riders, necessitating adjustments to the ride's frequency.

PREREQUISITES
  • Understanding of centripetal force and motion
  • Familiarity with static and kinetic friction equations
  • Knowledge of basic physics formulas, including F = ma and weight calculations
  • Ability to perform unit conversions, particularly for revolutions per minute to radians per second
NEXT STEPS
  • Calculate the normal force for various masses on a rotating ride
  • Explore the relationship between frequency and centripetal acceleration
  • Investigate safety standards for amusement park rides regarding friction and normal forces
  • Learn about the implications of varying ride dimensions on required friction coefficients
USEFUL FOR

Physics students, mechanical engineers, amusement ride designers, and anyone interested in the dynamics of rotational motion and safety in amusement park rides.

Pyrowolf
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Homework Statement


In a classic carnival ride, patrons stand against the wall in a cylindrically shaped room. Once the room gets spinning fast enough, the floor drops from the bottom of the room! Friction between the walls of the room and the people on the ride make them the “stick” to the wall so they do not slide down. In one ride, the radius of the cylindrical room is R = 6.3 m and the room spins with a frequency of 23.5 revolutions per minute.
1. Speed of the rider. I used circumfrence * revolutions/ 60 to get 15.5. (Already known to be correct)
2. What is the normal Force?
3. What is the minimum coefficient of friction needed between the wall and the person?
4)If a new person with mass 92 kg rides the ride, what minimum coefficient of friction between the wall and the person would be needed?
6) To be safe, the engineers making the ride want to be sure the normal force does not exceed 1.7 times each persons weight - and therefore adjust the frequency of revolution accordingly. What is the minimum coefficient of friction now needed?

Homework Equations


F=m*a
Fs= (meu)s*N (Static friction)
Fk = (meu)k*N (Kinetic Friction)
Circumfrence = 2*Pi*r
Weight = m*g

The Attempt at a Solution


1. I already did

2. Well Weight is down and friction is up. (It's a circle on the horizontal for a ride). The Normal force points inward of the circle. With no force pointing outward (I think?) So I tried N= mv^2/r which came up wrong. So I'm not sure what I'm missing here.

3. Once I know the N, I can calculate the friction as follows: weight = friction (no movement in the Y axis) so mg = (meu) N -> (meu) = mg/N (this is my thought process anyway)

4. Same as problem 3, just a different mass

6. Change N to be 92*1.7 then solve like 3 and 4.

Also I get no feedback on 4 and 6 so I can't know if I'm right or not.
 
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Pyrowolf said:
2. Well Weight is down and friction is up. (It's a circle on the horizontal for a ride). The Normal force points inward of the circle. With no force pointing outward (I think?) So I tried N= mv^2/r which came up wrong. So I'm not sure what I'm missing here.
N= mv^2/r is correct. Why do you think it's wrong? (Some online system?)
 
Hmm, I just redid the calculations and for some reason I got a different answer this time. Yes it's an online system. And after redoing the calculations I got the right answer. I must have missed a decimal, or number, or parenthesis. Thanks for the help.
 

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